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3 years ago

I need help with solving 378+16 but it is adding using groups of 10 and 100

Mathematics

1 answer:

hammer [34]3 years ago

7 0

394 is the answer using addition

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Help please, need help

iren2701 [21]

Answer:

Step-by-step explanation:

B) Small triangle:

height = 20 - 15 = 5

base = 1

height to base = 5 :1

Medium triangle :

height = 60 -30 = 30

base = 2

height to base = 30 :2 = 15 :1

Large triangle:

height = 120 - 60 = 60

base = 4

height to base = 60 : 4 = 15 : 1

c) The ratios are same. This is true because wage and working hours are in direct proportion.

E) Take any two points. (1 , 15) & (2 , 30). Find the slope.

Slope = Change in y ÷ change in x

m = (30-15) ÷ (2 -1) = 15/1 = 15

y -y1 = m(x -x1)

y - 15 = 15(x - 1)

y - 15 = 15x - 15

y = 15x - 15 + 15

y = 15x

6 0

3 years ago

Read 2 more answers

Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,

rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t , y_2(t) = 0 \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1$ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$